# Question 74eaa

Feb 12, 2016

$x \to {90}^{+} + 2 k \cdot {180}^{\circ}$ or $x \to {90}^{-}$$+ 2 k \cdot {180}^{\circ}$,

$x = {30}^{\circ} + 2 k \cdot {360}^{\circ}$ and $x = {150}^{\circ} + 2 k \cdot {360}^{\circ}$, where $k \in \mathbb{N}$

#### Explanation:

Substituting

$\cos 2 x = {\cos}^{2} x - {\sin}^{2} x = 2 {\cos}^{2} x - 1$

And

$\sin 2 x = 2 \sin x \cdot \cos x$

The expression becomes

$\rightarrow 1 - 2 {\cos}^{2} x + 1 + \frac{\sin \frac{x}{\cos} x}{1 - \sin \frac{x}{\cos} x} = 1 + 2 \sin x \cdot \cos x$

rarr2-2cos^2 x+(sinx/cancel(cosx))(cancel(cosx)/(cosx-sin x))-1- 2sinx*cosx=0

$\rightarrow 2 {\sin}^{2} x \left(\cos x - \sin x\right) + \sin x - \left(1 + 2 \sin x \cdot \cos x\right) \left(\cos x - \sin x\right) = 0$

rarr2sin^2 x*cosx-2sin^3 x+sin x -cosx+sinx-2sinx*cos^2 x+2sin^2 x*cosx=0

$\rightarrow 4 {\sin}^{2} x \cdot \cos x - 2 {\sin}^{3} x + 2 \sin x - \cos x - 2 \sin x \cdot {\cos}^{2} x = 0$

$\rightarrow 4 {\sin}^{2} x \cdot \cos x - 2 {\sin}^{3} x + 2 \sin x - \cos x - 2 \sin x \cdot \left(1 - {\sin}^{2} x\right) = 0$

rarr4sin^2 x*cosx-cancel(2sin^3 x)+cancel(2sinx)-cosx- cancel(2sinx)+cancel(2sin^3 x)=0#

$\rightarrow 4 {\sin}^{2} x \cdot \cos x - \cos x = 0$

$\rightarrow \cos x \left(4 {\sin}^{2} x - 1\right) = 0$

$\cos x = 0$ => $x = {90}^{\circ} + k \cdot {180}^{\circ} \to$ but in the original expression there's $t g x$ and $t g x$ is undefined for $x = {90}^{\circ} \mathmr{and} x = {270}^{\circ}$ so we must outrightly reject this solution or only admit it when $x \to {90}^{\circ}$

$4 {\sin}^{2} x - 1 = 0$ $\implies$ ${\sin}^{2} x = \frac{1}{4}$ $\implies$ $\sin x = \frac{1}{2}$ $\implies$ $x = {30}^{\circ} + k \cdot {360}^{\circ}$ or $x = {150}^{\circ} + k \cdot {360}^{\circ}$, where $k \in \mathbb{N}$